Algebra II Midterm Exam
A free Algebra II lesson from the “Rational Exponents and Radicals” unit, with a worked example and practice problems including step-by-step solutions.
This midterm exam covers the first half of Algebra II: functions (notation, domain/range, inverses, composition, piecewise), polynomial arithmetic, complex numbers, polynomial factorization and division, polynomial graphs, and radicals and rational exponents.
What you'll learn
- Demonstrate mastery of the first half of Algebra II
- Move fluently across functions, polynomials, complex numbers, factoring, division, graphs, and radicals
- Choose the right method on mixed problems
Worked example
Problem. Solve x^2 - 5x + 6 = 0 by factoring.
- Find two numbers that multiply to 6 and add to -5.
- -2 and -3 work, so (x - 2)(x - 3) = 0.
- Therefore x = 2 or x = 3.
Answer: x = 2 or x = 3
Lesson preview: this page shows 5 practice items. The interactive activity contains 45 practice items and 45 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Given f(x) = 3x + 2, find f(3).
Show solution
- Substitute x = 3.
- f(3) = 3(3) + 2 = 9 + 2.
- f(3) = 11.
Answer: 11
2. Find the domain of f(x) = 1/(x - 3). Give your answer as the single x-value that must be excluded.
Show solution
- The denominator x - 3 cannot equal 0.
- Set x - 3 = 0, so x = 3.
- Every real number except 3 is allowed, so x = 3 is excluded.
Answer: 3
3. If f(x) = x + 2, find a rule for the inverse f^-1(x).
Show solution
- Write y = x + 2, then swap x and y: x = y + 2.
- Solve for y: y = x - 2.
- So f^-1(x) = x - 2.
Answer: x - 2
4. Let f(x) = 3x + 2 and g(x) = x + 3. Find f(g(3)).
Show solution
- First evaluate the inside: g(3) = 3 + 3 = 6.
- Now apply f to that result: f(6) = 3(6) + 2 = 18 + 2.
- So f(g(3)) = 20.
Answer: 20
5. For f(x) = { 3x + 2, if x < 4; 6x + 8, if x ≥ 4 }, find f(-3).
Show solution
- Since -3 < 4, use the rule 3x + 2.
- 3(-3) + 2 = -9 + 2 = -7.
- So f(-3) = -7.
Answer: -7
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